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Absolute Value Function

|x2-x-6x-5/x2...

Question

If $∣ ∣ ∣ \frac{(x^{2} - x - 6) (x - 5)}{(x^{2} + 1) (x - 4)} ∣ ∣ ∣ = - \frac{(x^{2} - x - 6) (x - 5)}{(x^{2} + 1) (x - 4)}$ , then $x$ lies in

A

x \in [- 2, 3] \cup [4, \infty)

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B

x \in [- 2, 3] \cup (4, 5]

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C

x \in (- \infty, 3] \cup (4, 5]

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D

x \in [- 2, 5]

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Solution

The correct option is B $x \in [- 2, 3] \cup (4, 5]$
$∣ ∣ ∣ \frac{(x^{2} - x - 6) (x - 5)}{(x^{2} + 1) (x - 4)} ∣ ∣ ∣ = - \frac{(x^{2} - x - 6) (x - 5)}{(x^{2} + 1) (x - 4)}$
Above condition is possible only when
$\frac{(x^{2} - x - 6) (x - 5)}{(x^{2} + 1) (x - 4)} \leq 0$
as $x^{2} + 1$ is always positive
By wavy curve method
$\frac{(x - 3) (x + 2) (x - 5)}{(x^{2} + 1) (x - 4)} \leq 0$
$x \in [- 2, 3] \cup (4, 5]$

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Similar questions

Q. Largest interval in which the inequality

∣ ∣ ∣ \frac{(x + 3) (2 + x)}{(x - 1) (x - 5)} ∣ ∣ ∣ > \frac{x^{2} + 5 x + 6}{x^{2} - 6 x + 5}

Q. Which of the following are quadratic equations?

(i) x² + 6x − 4 = 0
(ii)

\sqrt{3} x^{2} - 2 x + \frac{1}{2} = 0

x^{2} + \frac{1}{x^{2}} = 5

x - \frac{3}{x} = x^{2}

2 x^{2} - \sqrt{3 x} + 9 = 0

x^{2} - 2 x - \sqrt{x} - 5 = 0

(vii) 3x² − 5x + 9 = x² − 7x + 3
(viii)

x + \frac{1}{x} = 1

(ix) x² − 3x = 0
(x)

{(x + \frac{1}{x})}^{2} = 3 (x + \frac{1}{x}) + 4

(xi) (2x + 1) (3x + 2) = 6(x − 1) (x − 2)
(xii)

x + \frac{1}{x} = x^{2}, x \neq 0

(xiii) 16x² − 3 = (2x + 5) (5x − 3)
(xiv) (x + 2)³ = x³ − 4
(xv) x(x + 1) + 8 = (x + 2) (x − 2)

x^{2} + 6 x - 27 > 0

- x^{2} + 3 x + 4 > 0

x

lies in the interval

Q. The factors of x³ − 7x + 6 are

(a) x (x − 6) (x − 1)

(b) (x² − 6) (x − 1)

(c) (x + 1) (x + 2) (x + 3)

(d) (x − 1) (x + 3) (x − 2)

Q. Which of the following are quadratic equation in x ?
(i)

x^{2} - x + 3 = 0

2 x^{2} + \frac{5}{2} x - \sqrt{3} = 0

\sqrt{2} x^{2} + 7 x + 5 \sqrt{2} = 0

\frac{1}{3} x^{2} + \frac{1}{5} x - 2 = 0

x^{2} - 3 x - \sqrt{x} + 4 = 0

x - \frac{6}{x} = 3

x + \frac{2}{x} = x^{2}

x^{2} - \frac{1}{x^{2}} = 5

{(x + 2)}^{3} = x^{3} - 8

(2 x + 3) (3 x + 2) = 6 (x - 1) (x - 2)

{(x + \frac{1}{x})}^{2} = 2 (x + \frac{1}{x}) + 3

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Absolute Value Function

Standard XII Mathematics

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